Convergence in $L^{p_1}$ and $L^{p_2}$

Question

Suppose $f_k$ is a sequence of $\mu$-measurable function. Let $p_1$ and $p_2\in[1,\infty)$, and $f_k\in L^{p_1}\cap L^{p_2}$. Also suppose that there exists $g\in L^{p_1}$ and $h\in L^{p_2}$ such that $f_k\to g$ in $L_{p_1}$, $f_k\to h$ in $L^{p_2}$. Prove that $g=h$ ($\mu$-a.e.).

Answer 1

Pick a sequence $n_k\uparrow \infty$ such that for each $k$, $\mu\{x,|f_{n_k}(x)-g(x)|\geqslant k^{-1}\}\leqslant 2^{-k}$: we have $f_{n_k}\to g$ almost everywhere.

Taking in a similar way $m_k\uparrow \infty$, $\{m_k,k\geqslant 1\}\subset\{n_k,k\geqslant 1\}$ with for each $k$, $\mu\{x,|f_{m_k}(x)-h(x)|\geqslant k^{-1}\}\leqslant 2^{-k}$, we get $g=h$ almost everywhere.